The ability of a material to transfer heat or electric current is called

Can anyone explain what weak force could do? I’ve been reading comics and I don’t understand what they can do when it says “they

Morgarella [4.7K]

Explanation:

To my understanding of comic books, weak force indicates that the hero was strong enough to overcome its powers...

4 0

4 years ago

Could you help me with a science question really quick?

Rama09 [41]

Yea I would love to help u

4 0

3 years ago

What do we know about the relative strength of the electric field around each object?

Vsevolod [243]

Answer:

A charged object in an electric field experiences a force due to the field. The electric field strength, E, at a point in the field is defined as the force per unit charge on a positive test charge placed at that point.

Explanation:

6 0

3 years ago

All of these are types of electromagnetic waves except for:

dimaraw [331]

C- Sigma Waves, This is what my teacher used to get us to remember https://www.youtube.com/watch?v=bjOGNVH3D4Y

3 0

3 years ago

A cannon fires a cannonball forward at a velocity of 48.1 m/s horizontally. If the cannon is on a mounted wagon 1.5 m tall, how

GalinKa [24]

Answer: 473.640 m

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the cannonball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=48.1 m/s$ is the cannonball's initial velocity

$\theta=0$ because we are told the cannonball is shot horizontally

$t$ is the time since the cannonball is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the cannonball

$y=0$ is the final height of the cannonball (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

We need to find how far (horizontally) the cannonball has traveled before landing. This means we need to find the maximum distance in the x-component, let's call it $X_{max}$ and this occurs when $y=0$ .

So, firstly we will find the time with (2):

$0=1.5 m+48. 1 m/s sin(0\°) t-(4.9 m/s^{2})t^2$ (3)

Rearranging the equation:

$0=-(4.9 m/s^{2})t^2+48. 1 m/s sin(0\°) t+1.5 m$ (4)

$-(4.9 m/s^{2})t^2+(48. 1 m/s) t+1.5 m=0$ (5)

This is a <u>quadratic equation</u> (also called <u>equation of the second degree</u>) of the form $at^{2}+bt+c=0$ , which can be solved with the following formula: